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<@&435155896780324864> Does anyone know how to do a Big O time complexity analysis on a Java function?

@Jacob example function?

```java
public static void two(int n)
{
if(n > 0)
{
System.out.println("n: " +n);
two(n - 1);
two(n - 1);
}
else if (n < 0)
{
two(n + 1);
two(n + 1);
System.out.println(″n: ″ + n);
}
}
```

It's mostly like doing it on paper, you have most of the same assumptions of the Random Access Machine

Do you have any skepticism that assumptions of RAM don't hold?

No, it's just a basic analysis. I need to provide a "logical justification" for my answer. Not sure if that means I have to write a paragraph or there's some kind of equation I need to use.

There's an equation.

Not necessarily expecting an answer to the question from anyone, but maybe something that can get me in the right direction

f(0) = O(1)

f(|n|) = 2*f(|n|-1) + O(1)

😬

so f(n) = O(2^n)

Is that the equation for this particular function?

yep

How would I go about getting that?

So start by looking at the function to understand what is the "base case". The base case is the input that doesn't cause a recursive call

ah, okay
and write the base case as one of those formulas?

the only input that doesn't have a recursive call is n=0

yep

so how do I move up from there?

So after that you really have two cases, one when n is > 0 and one when , is < 0. If you look closely what you have in each case is that n decreases or increases towards 0 with steps of 1. I'm just here pointing out the two cases. For simplicity consider what happens in just the positive case first

so for the positive case you have that f(n) involves a check on n, which is constant time, a printline which is constant time, and then two calls to f(n-1)

so in this n positive case
f(n) = the constant stuff + 2\*f(n-1) = O(1) + 2\*f(n-1)

But don't the calls to f(n-1) create even more calls?

since it's recursive

Yep

Do I need some kind of formula to determine how many calls it will create?

so by using the same formula, f(n-1) = 2*f(n-2) + o(1)

plugging that in to what you had before:
f(n) = 2\*(2\*f(n-2) + O(1)) + O(1) = 2\*2\*f(n-2) + O(1)

ah, so it just simplifies to another logarithmic function?

yes

its exponential (the inverse of logarithm)

Okay. That should be enough to write up a logical explanation, I think. Apparently the other two function are easier to analyze.

Thanks a lot

```java
public void three(int n)
{
int i, j, k;
for (i = n/2; i > 0; i = i/2)
for (j = 0; j < n; j++)
for (k = 0; k < n; k++)
System.out.println("i: " + i + " j: " + j+" k: " + k);
```

This is the next one. Should be easier for me to figure this one out since it's just for loops, right?

yes

```java
public static void four(int n)
{
if (n > 1)
{
System.out.println(n);
four(n-1);
}
for (int i = 0; i < n; i++)
System.out.println(i);
}
```

I'm pretty sure this one is just Nlog(N) but I might be wrong